# Coordinate speed of light vs. gravitational time dilation?

1. Jul 31, 2007

### kmarinas86

I know there is:

• A gravitational length contraction by the factor of $\sqrt{1-2GM/rc^2}$
• A time slowing by the factor of $\sqrt{1-2GM/rc^2}$

I would have thought the former and latter affect the notions of wavelength and frequency of light respectively, am I not right?

However, then we (the shruken and time-slowed) are left with a increase apparent wavelength of a distant photon by a factor of $1/\sqrt{1-2GM/rc^2}$ and a increase apparent frequency of a distant photon by a factor of $1/\sqrt{1-2GM/rc^2}$, resulting in a increase in the apparent speed of distant light by a factor of $1/(1-2GM/rc^2)$.

Then there is the coordinate speed of light which is reduced by the factor $1-2GM/rc^2$, where $r$ refers to the radius from the gravitating object where the light is considered.

But now we are stuck with another problem. We observers at $r$ will have a clock running a rate $\sqrt{1-2GM/rc^2}$ times as much than a distant observer not in orbit. But our coordinate speed of light at $r$ is $1-2GM/rc^2$ times as much-a larger difference! That would mean that the apparent speed of light for the distant observer would be $1/\sqrt{1-2GM/rc^2}$ times as much. ? Surely, I am understanding something wrong.

2. Jul 31, 2007

### kmarinas86

Apparently, in the last paragraph, I forgot to consider the effect of length contraction. The undoing of length contraction at a distance from the gravitational mass would have the effect of making the wavelength appear shorter, making the apparent speed of light the same-as it should be.

3. Jul 31, 2007

### meopemuk

Basically, I agree with you. Though I should warn you that my views on gravity are not "mainstream". Presently, they are being discussed at the "Independent research" subforum https://www.physicsforums.com/showthread.php?t=175965

Assuming the slowing down of light in the gravitational field by the factor $1-2GM/rc^2$ is the easiest way to explain the effect of Shapiro radar echo delay.

I am not sure how you made this conclusion. In my opinion, the speed of light at point $r$ is $c(1-2GM/rc^2)$ for all observers.

Regards.
Eugene.

Last edited: Jul 31, 2007
4. Jul 31, 2007

### pervect

Staff Emeritus
Using local clocks and rulers, the speed of light is always 'c'. For "coordinate speeds" there are a confusingly wide variety of available clocks and rulers that one might use. Using "local" clocks and rulers makes things simple - the speed of light at any location is always 'c', and time always ticks at 1 second per second.

I believe that the OP's question was basically how one reconciled the above with the well-known Schwarzschild coordinate chart.

Using the Schwarzschild line element

ds^2 = g_00 dt^2 + g_11 dr^2 + (angular terms)

We can say that:

proper time = sqrt(|g_00|) * $\delta t$

Here $\delta t$ represents a small change in the Schwarzschild time coordinate t, and $\delta r$ represents a small change in the Schwarzschild radial coordinate r.

We can say this because ds^2 = g_00 $(\delta t)^2$ if all the other delta's are zero. This implies that ds = sqrt(|g_00|) $\delta t$, where we take the absolute value of g_00 to insure that we are taking the square root of a positive quantity. Similarly

proper length = sqrt(|g_11|) * $\delta r$

Proper time and proper length are the lengths and times that one would measure in a local 'frame field', i.e. using local clocks and rulers, where the speed of light is always equal to 'c', and clocks read out proper time.

(Note that we are assuming here that our observer is stationary with respect to the large mass, i.e. he maintains a constant Schwarzschild r, $\theta$, and $\phi$ coordinates.)

This implies, using the well-known values for g_00 and g_11 in the Schwarzschild coordinate system, that

proper time = $\sqrt{1-2M/r} \, \delta t$
proper length = $\frac{1}{\sqrt{1-2M/r}} \, \delta r$

Based on this result, we can see that while the local speed of light stays constant at 'c', the coordinate speed of light decreases as one gets closer to a large mass, i.e. for a light beam, in geometric units

$$\frac{\delta r}{\delta t}$$ = 1 -2M/r

which approaches zero as one approaches the event horizon and approaches unity ('c') at infinity.

Last edited: Aug 1, 2007
5. Aug 26, 2007

### kmarinas86

Very interesting.

I have another question however.

If length contractions occur for objects subject to a gravitational field, do they also occur for objects in the early universe, where the scale factor was smaller?

Consider, however, that the coordinate speed of light is supposed to be proportional to the square of the length contraction factor. This coordinate speed of light, could logically be decomposed into the coordinate frequency and coordinate wavelength, as if it were. Under the normal situation, say for objects within our solar system, an observer "falling from infinity" would expect the coordinate wavelength to decreased by the square of the length contraction factor while the grounded observer is decreased size by that factor.

So the falling observer sees an decreased of wavelength relative to his/her self, eventhough he/she is shrinking!

I was never told that the length contraction was inversely proportional to scale factor of the universe. Actually its quite the opposite. The increase in the ratio of the wavelength of cosmic background radiation over the size of material objects in the universe is clearly in close correspondence to the proportion of the scale factor. That is, the light appears to increase its wavelength as the scale factor increases. However, this scale factor is normally associated with a redshift, which is the type found in General Relativity. How, then, can length contraction, that matches the Einstein Shift (i.e. redshift due to GR effects) in the deep universe, not be there? If it were there, then the size of objects would increase in proportion to the scale factor (scale factors is larger, right here, right now). But that means that the light and mass would expand proportionally. That is certainly false. There is no such length contraction that increases with redshift, as there is astronomical evidence that the cosmic background radiation interacts with galaxies that corresponds to the higher frequency of the background at the time, and the wavelengths of light are most essential when considering its interaction with matter. More evidence against that is that the objects would be exceedingly small if one were to look into the distant universe, galaxies of nearly identical structure would shrink in proportion their cosmological redshift, and we know that is false. You could imagine how "galaxies that expand proportionally to the scale factor" would have trouble fitting with the fact that angular diameter distance actually decreases with increasing redshifts beyond something like z=1.6 (or something like that).

We are now at a problem. The cosmological shift that does not have a length contraction in proportion to it, and also, the coordinate speed of light is supposed to be proportional to the square of the length contraction, and is therefore unaffected! You may ask, so what?

Here's the what:

(I) Apparent frequency of early-universe galaxy's light ..... (II) Apparent wavelength of early-universe galaxy's light

(I) Reduces¹ (II) Leaves alone --- Gravitational time dilation of distant object

+

(I) Leaves alone (II) Increases¹ --- difference

=

(I) Reduces¹ (II) Increases¹ --- final result

_____________________

(I) Apparent frequency of early-universe galaxy's light ..... (II) Apparent wavelength of early-universe galaxy's light

DIFFERENCE =

(I) Leaves alone (II) Decreases¹ --- Length contraction of early-universe galaxy

+

(I) Leaves alone (II) Increases² --- Increase in the coordinate speed of light

=

(I) Leaves alone (II) Increases¹ --- difference

¹ to the same degree as the length contraction
² to the same degree as the square of the length contraction

The problem we have here is the coordinate speed of light, which is not to be affected by the expansion of space (right?). If it is affected by it, please let me know, and I am wasting my time.

Now if it is not, then we have another option:

(I) Apparent frequency of early-galaxy's light ..... (II) Apparent wavelength of early-galaxy's light

DIFFERENCE =

(I) Leaves alone (II) Decreases¹ --- Length "anti-contraction" ??? of early-universe galaxy

=

(I) Leaves alone (II) Increases¹ --- difference

When we have this, we are assuming that galaxies are shrinking relative to wavelengths of the cosmic background radiation (that is astronomical fact, as stated earlier). But this does not go along with any change in the coordinate speed of light, which is not affected by the cosmological scale factor. So, while gravitational time dilation deals with the time aspect (frequency) of the radiation source (incoming low frequency microwave background radiation), the shrinking of galaxies(???) would deal with the relative shrinkage of galaxies with respect to the wavelengths of radiation. Why I say this is because it avoids any change in the coordinate speed of light, which is impossible in a vacuum of flat GR space ;).

Let's clarify this by letting me ask a few questions for you:

1) Is the coordinate speed of light affected by the scale factor?

2) If not, then the coordinate speed of light must be unaffected by the scale factor. If so, assuming we are right in that the frequency of cosmic background radiation decreases as the scale factor increases, the wavelength in relation to matter MUST increase. There is a mainstream mechanism proposed called the expansion of space. But cosmic inflation appears to be totally unrelated to any sort of gravitational length contraction! Where is cosmological length contraction, if any?

3) If there is not, is there any value in suggesting that expansion of space is relative, in that, the galactic observer sees the stellar observer in contracting space, while the stellar observer sees the galactic observer in expanding space?

4) As for the two views espoused in question #3, which view do you think is more "god-like" (just for fun) ;).

Last edited: Aug 26, 2007
6. Aug 26, 2007

### pervect

Staff Emeritus
What makes you say that? Consider -dt^2 + a(t)^2 * (dx^2 + dy^2 + dz^2), a flat FRW universe.

The coordinate speed of light is not proportional to a(t)^2. For instance, in the x direction

dt = a dx

is a null geodesici.e. a light beam, because the Lorentz interval is zero.

[edit-correction]

Thus for light, dx/dt = 1/a(t). This is an inverse linear relationship, a(t) does not get squared anywhere.

Given
dx = dt/a(t)

it's complex to solve in general, but for instance if a(t)=t then

x = ln(t) + k

Last edited: Aug 26, 2007
7. Aug 26, 2007

### kmarinas86

I was thinking "Schwarzschild" when talking about the squared, but then I guess I misapplied it to an expanding universe problem =P. My bad.

So in sum, in an expanding universe, the coordinate speed of light decreases in a way inversely related to the increasing scale factor? Oh great, does it depend on whether the rate of expansion varies over time? Or is this another one of those things that have multiple answers depending on how you do it?

8. Aug 27, 2007

### Jorrie

Is it correct to say that this is the radial coordinate speed of light, while the transverse coordinate speed of light in geometric units will be

$$r\omega = \sqrt{1 -2M/r}$$ ?

9. Aug 27, 2007

### pervect

Staff Emeritus
Well,
$$(-1 + 2M/r) dt^2 + r^2 d\theta^2 = 0$$

so if you assume $$\omega = \frac{d\theta}{dt}$$ and you consider the $\phi=0$ plane, you get the above equation. But usually people don't use $\omega$ for $d\theta / dt$. Note that sometimes $d\theta / d \tau$ might be more useful, i.e. if one were to compute the angular momentum.

See for instance http://www.fourmilab.ch/gravitation/orbits/, note however that they work problems with $\phi=0$ but with $\theta=0$. Also, they are more concerned with the orbits of massive particles than the "orbits" (geodesics) of light rays, though similar principles apply.

Writing out $d\theta / dt$ makes it clear with which time variable you are performing the derivative, i.e. Scharzschld coordinate time t, or proper time $\tau$.

I also generally encourarge people not to think of d (space coordinate) / d (time coordinate) as a "velocity", it doesn't really have the same physical properties.

Last edited: Aug 27, 2007
10. Aug 28, 2007

### Jorrie

Local vs. Schwarzschild coordinate velocities

Thanks pervect. I take note of:

If one restricts d(space coordinate) to an infinitesimal and well specified region and always use the correct d(time coordinate), would your discouragement soften?

I do find it rather useful to think about 'locally measured velocities' v_(loc) and 'Schwarzschild coordinate velocities' (v) which can be transformed from the one into the other via the two factors for coordinate radial and transverse velocity of light respectively.

So, I take it that for particles with mass, the transformations can be written in geometrical units as:

$$v_r = (1-2M/r) v_{r(loc)}$$ and $$v_t = \sqrt{(1-2M/r)} v_{t(loc)}$$

where v_r is the coordinate radial velocity, v_t the coordinate transverse velocity and r the Schwarzschild radial coordinate of the particle.

I suspect there are situations where this may lead to problems, but can't think of any outside of the event horizon of a Schwarzschild hole.

Jorrie

Last edited: Aug 28, 2007
11. Aug 29, 2007

### pervect

Staff Emeritus
Basically, to get the usual notion of velocity that one can safely apply one's intuition to, one needs to impose a local coordinate system so that the local metric is Minkowskian, i.e. the metric has the form

ds^2 = -dtt^2 + drr^2 (1)

or more generally, with multiple dimensions

ds^2 = -dtt^2 + dxx^2 + dyy^2 + dzz^2 (2)

where tt is a local time coordinate, and rr (or xx,yy,zz) are local distance coordinates.

In this case, d (local space coordinate) / d (local time coordinate) acts in the way that one expects. Because the region is small enough, curvature effects can be ignored. The only other thing that needs to be done is to use a standard orthonormal coordinate system, rather than a generalized one. Orthonormal means that one unit of the local time coordinate has a unit length, that one unit of the local space coordinate has a unit length, and that time and space are perpendicular. If one has a metric of the form given in (1), one is guaranteed that the local space and time coordinates rr and tt are orthonormal.

In the Schwarzschld metric, this process involves rescaling both the space and time coordinates.

A detailed example of this approach is done in an old post of mine
[correction]
https://www.physicsforums.com/showpost.php?p=602558&postcount=29

here rr is the local coordinate displacement representing distance, and tt is the local coordinate displacement representing time, from the viewpoint of a stationary observer near a black hole.

Note that one of the things that using the form of the metric 2 does is insure that the coordinate speeds of light are isotropic, as one physically expects.

We may be talking about totally different things when we talk about locally measured velocities, then. Note that my locally measured velocities will always be in a coordinate system where the speed of light is isotropic, and that this is a key point in applying one's intuition properly.

If I'm understanding your defintion of local velocities, it is equivalent to using inches (for example) to measure distance going in the radial direction, and feet in the transverse direction, and then you might say

"look, the speed of light is 12 times as big in the transverse direction as it is in the radial direction"

but it's just a consequence of the funky coordinates you've used, if you use standardized coordinates the speed of light will appear to be isotropic.

Last edited: Aug 30, 2007
12. Aug 29, 2007

### Jorrie

No, I'm in full agreement with your definition of local coordinates and velocities given above. No 'funky coordinates', I hope.

What I'm saying is that once you have measured such a local velocity (in local coordinates where the speed of light is isotropic and all that) you can take the local radial velocity component (radial relative to the Schwarzschild hole) and transform it to a Schwarzschild coordinate radial velocity by multiplying it by (1-2M/r). Same for the momentarily transverse local velocity component, just the factor is the square root of (1-2M/r), with r the Schwarzschild radial coordinate.

To be somewhat more precise, say I'm a momentarily stationary inertial observer at Schwarzschild radius r=4M, which is also the origin of my local inertial coordinate system (x,y,tau), oriented so that the x-axis ties up with the Schwarzschild radial direction and the y-axis is orthogonal to the radial (my transverse direction).

I measure the local radial and transverse velocity components of an object at the origin of my inertial coordinates, using my local clock and ruler (or my local radar) as: dx/dtau = 0.5 and dy/dtau = 0.5 (geometric velocities).

Now I transform those velocities to Schwarzschild coordinates (x',y',t) and get dx'/dt = (1-2/4) x 0.5 = 0.25 and dy'/dt = sqrt(1-2/4) x 0.5 = 0.3536.

If this is not valid, I'm missing something and would like to know what...

Last edited: Aug 30, 2007
13. Aug 30, 2007

### pervect

Staff Emeritus
OK, first a correction - the old post that I was talking about was:
https://www.physicsforums.com/showpost.php?p=602558&postcount=29

Let me try and work this through in detail:

At r=4M, 1-2M/r = 1 - 2M/4M = .5 and so

ds^2 = -.5 dt^2 + 2 dr^2 + (4M)^2 d phi^2=

- (.707 * dt)^2 + (1.414 * dr)^2 + (4M * dphi)^2

So the transformation tau = sqrt(1/2)*t and X = sqrt(2)*(r-4M) and Y = 4M * dphi

translates Schwarzschild coordinates (t,r,phi) in the theta=0 plane into local inertial coordinates (tau,X,Y)

X points in the radial direction, and Y points in the transverse direction.

To find the (physical) velocity dX/dtau, we would write:

dX/dtau = (dr/dtau) * (dX/dr)

we see from the above that dX/dr = sqrt(2), so

dx/dtau = sqrt(2) (dr/dtau)

or perhaps

dX/dtau = (dr/dt) * (dX/dr) * (dt/dtau)

dtau/dt = sqrt(1/2) and dt/dtau = sqrt(2)

so dX/dtau = 2 (dr/dt)

I don't think this quite matches up with what you wrote, but I hope it's enough for you to work with and resolve any discrepancies.

Similarly

dY/dtau = (dphi/dt) * (dY/dphi) * (dt/dtau), but I won't go into any more detail.

14. Aug 30, 2007

### Jorrie

Thanks pervect, but this gives exactly the numerical result that I got! (dx/dtau= 0.5, dr/dt=0.25)

I worked this through and got: dY/dtau = r(dphi/dt)*sqrt(2) for this case, in agreement with my numerical result of: dY/dtau = 0.5 and r(dphi/dt) = dy'/dt = 0.3536.

So I still hold that with proper care, the Schwarzschild coordinate velocities can be transformed to local velocities (and visa-versa) via the Schwarzschild radial and transverse velocities of light. Restating my post #10's equations in the terms that you used above:

Radial component: $$\frac{dr}{dt} = (1-2M/r) \ \frac{dX}{d\tau}$$

Transverse component: $$r\frac{d\phi}{dt} = \sqrt{1-2M/r} \ \frac{dY}{d\tau}$$

From my 'engineering view', this is very useful.

15. Aug 30, 2007

### pervect

Staff Emeritus
I couldn't quite figure out what you meant by "Schwarzschild coordinates" (x', y', t). I was looking for r, theta, and phi.

Anyway, I'm glad that we are in basic agreement once we get the terminology issues straightened out.